Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Sure, let's solve this question step-by-step.
### Given Data
- Temperature [tex]\( T = 25^{\circ}C \)[/tex]
- Frequency [tex]\( f = 512 \)[/tex] Hz
- Speed of sound at [tex]\( 0^{\circ}C \)[/tex], [tex]\( V_0 = 330 \, \text{m/s} \)[/tex]
### (i) Velocity of Sound at [tex]\( 25^{\circ}C \)[/tex]
The speed of sound in air increases by approximately [tex]\( 0.6 \, \text{m/s} \)[/tex] for each degree Celsius increase in temperature. Therefore, the speed of sound at [tex]\( 25^{\circ}C \)[/tex] can be calculated as follows:
[tex]\[ V_{25} = V_0 + (0.6 \times T) \][/tex]
Substituting the given values:
[tex]\[ V_{25} = 330 \, \text{m/s} + (0.6 \times 25) \, \text{m/s} \][/tex]
[tex]\[ V_{25} = 330 \, \text{m/s} + 15 \, \text{m/s} \][/tex]
[tex]\[ V_{25} = 345 \, \text{m/s} \][/tex]
So, the velocity of sound at [tex]\( 25^{\circ}C \)[/tex] is [tex]\( 345 \, \text{m/s} \)[/tex].
### (ii) Length of the Closed Tube at First Resonance
The first resonance in a closed tube occurs when the length of the tube is equal to one-quarter of the wavelength of the sound wave.
First, we need to find the wavelength ([tex]\( \lambda \)[/tex]) of the sound wave using the speed of sound and the frequency. The relation between wavelength, speed of sound, and frequency is given by:
[tex]\[ \lambda = \frac{V_{25}}{f} \][/tex]
Substituting the known values:
[tex]\[ \lambda = \frac{345 \, \text{m/s}}{512 \, \text{Hz}} \][/tex]
[tex]\[ \lambda = 0.673828125 \, \text{m} \][/tex]
The length of the tube at the first resonance corresponds to a quarter of the wavelength:
[tex]\[ \text{Length} = \frac{\lambda}{4} \][/tex]
Substituting the wavelength value:
[tex]\[ \text{Length} = \frac{0.673828125 \, \text{m}}{4} \][/tex]
[tex]\[ \text{Length} = 0.16845703125 \, \text{m} \][/tex]
So, the length of the closed tube at the first resonance is [tex]\( 0.16845703125 \, \text{m} \)[/tex].
### Summary
1. Velocity of sound at [tex]\( 25^{\circ}C \)[/tex]: [tex]\( 345 \, \text{m/s} \)[/tex]
2. Length of the closed tube at the first resonance neglecting end correction: [tex]\( 0.16845703125 \, \text{m} \)[/tex]
### Given Data
- Temperature [tex]\( T = 25^{\circ}C \)[/tex]
- Frequency [tex]\( f = 512 \)[/tex] Hz
- Speed of sound at [tex]\( 0^{\circ}C \)[/tex], [tex]\( V_0 = 330 \, \text{m/s} \)[/tex]
### (i) Velocity of Sound at [tex]\( 25^{\circ}C \)[/tex]
The speed of sound in air increases by approximately [tex]\( 0.6 \, \text{m/s} \)[/tex] for each degree Celsius increase in temperature. Therefore, the speed of sound at [tex]\( 25^{\circ}C \)[/tex] can be calculated as follows:
[tex]\[ V_{25} = V_0 + (0.6 \times T) \][/tex]
Substituting the given values:
[tex]\[ V_{25} = 330 \, \text{m/s} + (0.6 \times 25) \, \text{m/s} \][/tex]
[tex]\[ V_{25} = 330 \, \text{m/s} + 15 \, \text{m/s} \][/tex]
[tex]\[ V_{25} = 345 \, \text{m/s} \][/tex]
So, the velocity of sound at [tex]\( 25^{\circ}C \)[/tex] is [tex]\( 345 \, \text{m/s} \)[/tex].
### (ii) Length of the Closed Tube at First Resonance
The first resonance in a closed tube occurs when the length of the tube is equal to one-quarter of the wavelength of the sound wave.
First, we need to find the wavelength ([tex]\( \lambda \)[/tex]) of the sound wave using the speed of sound and the frequency. The relation between wavelength, speed of sound, and frequency is given by:
[tex]\[ \lambda = \frac{V_{25}}{f} \][/tex]
Substituting the known values:
[tex]\[ \lambda = \frac{345 \, \text{m/s}}{512 \, \text{Hz}} \][/tex]
[tex]\[ \lambda = 0.673828125 \, \text{m} \][/tex]
The length of the tube at the first resonance corresponds to a quarter of the wavelength:
[tex]\[ \text{Length} = \frac{\lambda}{4} \][/tex]
Substituting the wavelength value:
[tex]\[ \text{Length} = \frac{0.673828125 \, \text{m}}{4} \][/tex]
[tex]\[ \text{Length} = 0.16845703125 \, \text{m} \][/tex]
So, the length of the closed tube at the first resonance is [tex]\( 0.16845703125 \, \text{m} \)[/tex].
### Summary
1. Velocity of sound at [tex]\( 25^{\circ}C \)[/tex]: [tex]\( 345 \, \text{m/s} \)[/tex]
2. Length of the closed tube at the first resonance neglecting end correction: [tex]\( 0.16845703125 \, \text{m} \)[/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.